Generate random numbers - specify range and mean

[Chiara Piazzo]

Hello everyone,

Trust everyone is keeping safe. Apologies if this might be a silly question - is there a way generate a set of random numbers (e.g. 100) within a given range (e.g [0 30] but with a specific mean (e.g. (e.g. 8.5)? Any help will be appreciated. Thank you!

Best,

Chiara

[Joro Kolev]

The question is not silly, in fact it is rather complicated (and probably impossible to answer) at the level of generality you are asking it.

But if I take your examples given as what you want to achieve here it is how:

. set obs 1000000
number of observations (_N) was 0, now 1,000,000

. gen beta = 30*rbeta(1,2.52941176471)

. summ beta

    Variable |        Obs        Mean    Std. Dev.       Min        Max
-------------+---------------------------------------------------------
        beta |  1,000,000    8.500319    6.353481    .000027    29.8619

[Chiara Piazzo]

Thank you for your response and the suggestion. Could you kindly give me a bit more information on

rbeta(1,2.52941176471)

Best,

Chiara

[Joro Kolev]

I did the following:

1) You asked me for a random variable within a certain range, the example you gave was [0,30]. I know that the Beta distribution is between [0,1], but I can fix this by multiplying by 30, then I get your range.

2) I also know that the Mean(Beta) = a/(a+b). So I solved the equation 30*a/(a+b) = 8.5. There are multiple solutions to that, but if I fix a = 1, then b has to be = 2.52941176471

3) So I went and generated a Beta random variable with parameters a=1 and b=2.52941176471.

And viola (as the French say) I got a number which had the properties you gave as an example.

Thank you for your response and the suggestion. Could you kindly give me a bit more information on

rbeta(1,2.52941176471)

Best,

Chiara

[Nick Cox]

From the identity


8.5 = .(8.5/30) * 30 + (21.5/30) * 0


it follows that a distribution which is 30 8.5/30 of the time and 0 otherwise satisfies your criteria. The mean and the range don't fix the distribution in general.

But as seriously suggested and implemented by Joro Kolev, a beta distribution is certainly a flexible family which may be adequate for your purposes.

PS I think that the French (at least those in my school texts) say voilà. But then again my school texts told me that the French say Quelle horreur! when annoyed, and I think those texts lied to me. My sampling of friends and colleagues indicates that faced with (say)

invalid syntax
r(198);

the fraction of French users saying "Quelle horreur!" is precisely 0.

[Joro Kolev]

This is excellent, Nick, a two point distribution the way how you construct it does the job here, so in this context I over-complicated unnecessarily. It did not even come to me, because what I have done in my own research is having such two point distributions given, and then trying to come up with a continuous distributions with the same (or as close as possible) corresponding moments. (E.g., in the wild bootstrap there are those Mammen and Radamacher two point distributions, so then how do we come up with something that is continuous and has similar moments.)

PS.

The joke which I tried to crack makes an allusion to a clueless person who tries to look sophisticated by insinuating that he knows French and tries to insert voila, but because he is clueless he says violin, fiddle.

The joke that you constructed was pretty awesome. Indeed, what they teach us in school is not exactly what the French say when they hit their finger with a hammer.

From the identity


8.5 = .(8.5/30) * 30 + (21.5/30) * 0


it follows that a distribution which is 30 8.5/30 of the time and 0 otherwise satisfies your criteria. The mean and the range don't fix the distribution in general.

But as seriously suggested and implemented by Joro Kolev, a beta distribution is certainly a flexible family which may be adequate for your purposes.

PS. I think that the French (at least those in my school texts) say voilà. But then again my school texts told me that the French say Quelle horreur! when annoyed, and I think those texts lied to me. My sampling of friends and colleagues indicates that faced with (say)

invalid syntax
r(198);

the fraction of French users saying "Quelle horreur!" is precisely 0.

[Chiara Piazzo]

Dear Joro Kolev and Nick Cox,

Thank you so much! This is great - I am playing around with your suggestions.

I have a question on this - what if I wanted to expand the range of values and include negative numbers [-60 20]? How would things change? Thanks again!

Best,

Chiara

P.S. Nick - thanks for the excellent joke

[Joro Kolev]

The two point distribution that Nick described is easier to handle. So you solve for p

p*(-60) + (1-p)*20 = 8.5,
giving p = 0.14375.

clear 

set obs 1000000

gen x = cond(runiform()<0.14375, -60, 20)

summ x

. summ x

    Variable |        Obs        Mean    Std. Dev.       Min        Max
-------------+---------------------------------------------------------
           x |  1,000,000     8.50864    28.05812        -60         20

Dear Joro Kolev and Nick Cox,

Thank you so much! This is great - I am playing around with your suggestions.

I have a question on this - what if I wanted to expand the range of values and include negative numbers [-60 20]? How would things change? Thanks again!

Best,

Chiara

P.S. Nick - thanks for the excellent joke

[Chiara Piazzo]

This is great, thanks again!

Best

CP

[Chiara Piazzo]

Sorry spoke too soon. The following code only generates values of -60 and 20. Is this what is supposed to happen?

gen x = cond(runiform()<0.14375, -60, 20)

Thanks and best.

Chiara

[Joro Kolev]

Yes, it is a two point distributing where the value of -60 happens with probability of 0.14375, and the value of 20 with probability (1-0.14375). It has a range of -60 to 20, and a mean of 8.5 as you asked.

Sorry spoke too soon. The following code only generates values of -60 and 20. Is this what is supposed to happen?

gen x = cond(runiform()<0.14375, -60, 20)

Thanks and best.

Chiara

[Chiara Piazzo]

Thank you!